Subspace.md

Linear SpacesSubspacesDefinition: Let VVV be a vector space. A subset WWW of VVV is called a subspace of VVV iff WWW is a vector space with respect to the operations of VVV, that is,
(S1) w1+w2∈Ww_1 + w_2 \in Ww1​+w2​∈W   ∀w1,w2∈W\forall w_1, w_2 \in W∀w1​,w2​∈W (closure under addition)
(S2) cw∈Wc w \in Wcw∈W   ∀c∈F\forall c \in F∀c∈F and ∀w∈W\forall w \in W∀w∈W (closure under scalar multiplication)
Remark: WWW is a subspace of VVV iff WWW is nonempty and WWW is closed under addition and scalar multiplication. All other axioms are inherited from the original vector space VVV.
Example: linear space V=R2V = \mathbb{R}^2V=R2, subspace W=[α 0]T∈R2:α∈RW = [\alpha \space 0]^T \in \mathbb{R}^2 : \alpha \in \mathbb{R}W=[α 0]T∈R2:α∈R

Solution: Let  w1=[α1 0]Tw_1 = [\alpha_1 \space 0]^Tw1​=[α1​ 0]T and w2=[α2 0]Tw_2 = [\alpha_2 \space 0]^Tw2​=[α2​ 0]T and c∈Rc \in \mathbb{R}c∈R,
(S1) w1=[α10]w_1 = \begin{bmatrix} \alpha_1 \\ 0 \end{bmatrix}w1​=[α1​0​] and w2=[α20]w_2 = \begin{bmatrix} \alpha_2 \\ 0 \end{bmatrix}w2​=[α2​0​] be two arbitrary elements of WWW. Then w1+w2=[α1+α20]∈Ww_1 + w_2 = \begin{bmatrix} \alpha_1 + \alpha_2 \\ 0 \end{bmatrix} \in Ww1​+w2​=[α1​+α2​0​]∈W
(S2) cw1=[cα10]∈Wc w1 = \begin{bmatrix} c \alpha_1 \\ 0 \end{bmatrix} \in Wcw1=[cα1​0​]∈W for all c∈Rc \in \mathbb{R}c∈R.
Hence WWW is a subspace of VVV.
Example: linear space V=R2V = \mathbb{R}^2V=R2, subspace W=[α 1]T∈R2:α∈RW = [\alpha \space 1]^T \in \mathbb{R}^2 : \alpha \in \mathbb{R}W=[α 1]T∈R2:α∈R

Solution: Let  w1=[α1 1]Tw_1 = [\alpha_1 \space 1]^Tw1​=[α1​ 1]T and w2=[α2 1]Tw_2 = [\alpha_2 \space 1]^Tw2​=[α2​ 1]T
(S1) w1=[α11]w_1 = \begin{bmatrix} \alpha_1 \\ 1 \end{bmatrix}w1​=[α1​1​] and w2=[α21]w_2 = \begin{bmatrix} \alpha_2 \\ 1 \end{bmatrix}w2​=[α2​1​] be two arbitrary elements of WWW. Then w1+w2=[α1+α22]∉Ww_1 + w_2 = \begin{bmatrix} \alpha_1 + \alpha_2 \\ 2 \end{bmatrix} \notin Ww1​+w2​=[α1​+α2​2​]∈/W
Remark: In R2\mathbb{R}^2R2, a subspace is a line through the origin. Any line through the origin is a subspace of R2\mathbb{R}^2R2.
In function spaces, an example can be given as follows:

Example: linear space V=V =V= set of all real-valued functions of a real variable t→f(t)t\rightarrow f(t)t→f(t);

subspace W1=W_1 =W1​= set of all continuous functions [+]

subspace W2=W_2 =W2​= set of all constant functions  [+]

subspace W3=W_3 =W3​= set of all functions periodic with π\piπ [+]
subspace W4=W_4 =W4​= set of all functions which are discontinuous at t=1t=1t=1 [-]
Remark: 0 vector is a subspace of any vector space, even subspace of itself, and it is the smallest subspace.
[+] W1,W2,W3W_1, W_2, W_3W1​,W2​,W3​ are subspaces of VVV

 W4W_4W4​ is not a subspace of VVV
Example:  Show that Y+ZY+ZY+Z is a linear subspace of XXX, if YYY and ZZZ are also linear subspaces of XXX.

Proof: Let w1+w2∈Ww_1+w_2 \in Ww1​+w2​∈W, with

w1=y1+z1w_1=y_1+z_1w1​=y1​+z1​ where y1∈Y, z1∈Zy_1 \in Y, \ z_1 \in Zy1​∈Y, z1​∈Z

w2=y2+z2w_2=y_2+z_2w2​=y2​+z2​ where y2∈Y, z2∈Zy_2 \in Y, \ z_2 \in Zy2​∈Y, z2​∈Z

then
w1+w2=y1+z1+y2+z2=(y1+y2)+(z1+z2)y1+y2∈Y,z1+z2∈Z⇒w1+w2∈Y+Z\begin{align*}
w_1+w_2=y_1+z_1+y_2+z_2 &=(y_1+y_2)+(z_1+z_2) \\
y_1+y_2 \in Y , z_1+z_2 \in Z &\Rightarrow w_1+w_2 \in Y+Z
\end{align*}w1​+w2​=y1​+z1​+y2​+z2​y1​+y2​∈Y,z1​+z2​∈Z​=(y1​+y2​)+(z1​+z2​)⇒w1​+w2​∈Y+Z​Shows that Y+ZY+ZY+Z is closed under addition. (lemma 1)
Let cw1∈Wc w_1 \in Wcw1​∈W, ∀c∈F\forall c \in F∀c∈F
cw1=c(y1+z1)=(cy1)+(cz1)cy1∈Y,cz1∈Z⇒cw1∈Y+Z\begin{align*}
c w_1=c(y_1+z_1) &=(cy_1)+(cz_1) \\
cy_1 \in Y , cz_1 \in Z &\Rightarrow cw_1 \in Y+Z
\end{align*}cw1​=c(y1​+z1​)cy1​∈Y,cz1​∈Z​=(cy1​)+(cz1​)⇒cw1​∈Y+Z​Shows that Y+ZY+ZY+Z is closed under scalar multiplication. (lemma 2)
Hence Y+ZY+ZY+Z is a linear subspace of XXX.
Example:  If YYY and ZZZ are subspaces of X, then Y∩ZY \cap ZY∩Z is a subspace of XXX.

 ~ Proof:
0∈Y0 \in Y0∈Y, 0∈Z0 \in Z0∈Z then by definition 0∈Y∩Z0 \in Y \cap Z0∈Y∩Z
for u,w∈Yu,w \in Yu,w∈Y and  u,w∈Z  ⟹  u,w∈Y∩Zu,w \in Z \implies u,w \in Y \cap Zu,w∈Z⟹u,w∈Y∩Z
now we need to show that u+w∈Y∩Zu+w \in Y \cap Zu+w∈Y∩Z (closure under addition)
starting  u∈Y,w∈Y  ⟹  u+w∈Yu \in Y, w \in Y \implies u+w \in Yu∈Y,w∈Y⟹u+w∈Y
similarly u∈Z,w∈Z  ⟹  u+w∈Zu \in Z, w \in Z \implies u+w \in Zu∈Z,w∈Z⟹u+w∈Z
hence u+w∈Y∩Zu+w \in Y \cap Zu+w∈Y∩Z

now we need to show that cu∈Y∩Zcu \in Y \cap Zcu∈Y∩Z (closure under scalar multiplication)
starting  u∈Y  ⟹  cu∈Yu \in Y \implies cu \in Yu∈Y⟹cu∈Y, ∀c∈F\forall c \in F∀c∈F
similarly u∈Z  ⟹  cu∈Zu \in Z \implies cu \in Zu∈Z⟹cu∈Z, ∀c∈F\forall c \in F∀c∈F
hence cu∈Y∩Zcu \in Y \cap Zcu∈Y∩Z

Hence Y∩ZY \cap ZY∩Z is a subspace of XXX.
Example: For YYY and ZZZ are subspaces of XXX, show that whether Y∪ZY \cup ZY∪Z is a subspace of XXX or not.

 ~ Proof:
Prove by contradiction.
Assume Y∪ZY \cup ZY∪Z is a subspace of XXX. Then Y∪ZY \cup ZY∪Z is closed under addition and scalar multiplication.
Let Y={(y,0):y∈R}Y =\{(y,0):y \in \mathbb{R}\}Y={(y,0):y∈R} and Z={(0,z):z∈R}Z =\{(0,z):z \in \mathbb{R}\}Z={(0,z):z∈R}.
Then u1=(1,0)∈Yu_1=(1,0) \in Yu1​=(1,0)∈Y and u2=(0,1)∈Zu_2=(0,1) \in Zu2​=(0,1)∈Z.
u1+u2=(1,1)∉Y∪Zu_1+u_2=(1,1) \notin Y \cup Zu1​+u2​=(1,1)∈/Y∪Z.
Example: Is R2\mathbb{R}^2R2 a subspace of complex vector space C2\mathbb{C}^2C2?

 ~ Proof: Note that we consider complex vector space, so if x∈R2x \in \mathbb{R}^2x∈R2 then x∈C2x \in \mathbb{C}^2x∈C2.
i∈Ci \in \mathbb{C}i∈C and x∈R2x \in \mathbb{R}^2x∈R2 then ix∈R2ix \in \mathbb{R}^2ix∈R2.
i(1,1)=(i,i)∉R2i(1,1) =(i,i) \notin \mathbb{R^2}i(1,1)=(i,i)∈/R2

Hence R2\mathbb{R}^2R2 is not a subspace of C2\mathbb{C}^2C2.
 ~ 
SubspacesSums of SubspacesDefinition: Suppose W1,...,WmW_1,...,W_mW1​,...,Wm​ are subspaces of a vector space VVV. The sum of W1,...,WmW_1,...,W_mW1​,...,Wm​ is the set of all possible sums of elements of W1,...,WmW_1,...,W_mW1​,...,Wm​:
W1+...+Wm={w1+...+wm:wi∈Wi,i=1,...,m}W_1 + ... + W_m = \{w_1 + ... + w_m : w_i \in W_i, i = 1,...,m\}W1​+...+Wm​={w1​+...+wm​:wi​∈Wi​,i=1,...,m}Remark: W1+...+WmW_1 + ... + W_mW1​+...+Wm​ is the smallest subspace of VVV containing W1,...,WmW_1,...,W_mW1​,...,Wm​.
Direct SumsDefinition: Suppose W1,...,WmW_1,...,W_mW1​,...,Wm​ are subspaces of a vector space VVV. The sum W1+...+WmW_1 + ... + W_mW1​+...+Wm​ is called a direct sum if each element of W1+...+WmW_1 + ... + W_mW1​+...+Wm​ can be written in one and only one way as a sum w1+...+wmw_1 + ... + w_mw1​+...+wm​ with wi∈Wiw_i \in W_iwi​∈Wi​.
Remark: if W1+...+WmW_1+...+W_mW1​+...+Wm​ is a direct sum, then W1⊕...⊕WmW_1 \oplus ... \oplus W_mW1​⊕...⊕Wm​ is used to denote the direct sum.
Example: Suppose UjU_jUj​ is a subspace of FnF^nFn of those vectors whose jjjth component is zero. As U2={(0,x,...,0)∈Fn:x∈F}U_2 = \{(0,x,..., 0) \in F^n : x \in F\}U2​={(0,x,...,0)∈Fn:x∈F}, Then.
U1⊕...⊕Un=FnU_1 \oplus ... \oplus U_n= F^nU1​⊕...⊕Un​=FnCondition for Direct Sum: Suppose W1,...,WmW_1,...,W_mW1​,...,Wm​ are subspaces of a vector space VVV. Then W1+...+WmW_1 + ... + W_mW1​+...+Wm​ is a direct sum iff the only way to write 0 as a sum w1+...+wmw_1 + ... + w_mw1​+...+wm​ with wi∈Wiw_i \in W_iwi​∈Wi​ is by taking w1=...=wm=0w_1 = ... = w_m = 0w1​=...=wm​=0. In other words, Suppose WWW and UUU are subspaces of a vector space VVV. Then W⊕UW \oplus UW⊕U is a direct sum iff W∩U={0}W \cap U = \{0\}W∩U={0}.
Proof: Proving the statements with iff is equivalent to proving the statements separately, in both directions.
W⊕U is a direct sum  ⟺  W∩U={0}W \oplus U \text{ is a direct sum} \iff W \cap U = \{0\}W⊕U is a direct sum⟺W∩U={0}